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a particle is projected at an angle tita to the horizontal with an initial velocity U.if the maximum horizontal distance travelled is 20m and the greatest height reached is 10m.find U and Tita

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  1. the horizontal distance is

    x(t) = u cosθ t
    The time taken to go 20m is thus
    t = 20/(u cosθ)

    The trajectory is
    y(t) = u sinθ t - g/2 t^2
    We know that y=0 at 0 and 20/(u cosθ), so the max height is reached at 10/(u cosθ). So,

    u sinθ * 10/(u cosθ) - g/2 (100/(u^2 cos^2 θ) = 10

    y = tanθ x - g/(2 (u cosθ)^2) x^2

    We know that y(20) = 0 and y(10) = 10. So, approximating g as 5 m/s^2,

    20tanθ - 500/(u cosθ)^2 = 0
    10tanθ - 125/(u cosθ)^2 = 10

    u = 5/2 √10
    tanθ = (√5-1)/2

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    posted by Steve
  2. Initial vertical speed Vi = U sin theta
    constant horizontal speed = u = U cos theta

    horizontal problem
    20 = U cos theta * T where T is time in air

    vertical problem
    v = Vi - g t
    v = 0 at the top
    so
    t at top = Vi/g
    g = 9.8 m/s^2 on earth
    so
    t at top = Vi/9.8

    now at the top at 10 meters
    h = 0 + Vi t -4.9 t^2
    10 = Vi t - 4.9 t^2
    10 = Vi^2/9.8 - 4.9 Vi^2/96
    10 = Vi^2 (.051)
    Vi = 14 m/s
    t at top = 14/9.8 = 1.43 s
    total time in air = 2t = 4.29 s
    20 = u (4.29)
    so u = 4.67 m/s
    so
    U = sqrt (4.67^2+14^2) = 14.8 m/s
    and
    tan theta = 14/4.67
    theta = 71.6 degrees

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    posted by Damon
  3. t at top = 14/9.8 = 1.43 s
    total time in air = 2t = 2.86 s
    20 = u (2.86)
    so u = 6.99 m/s
    so
    U = sqrt (6.99^2+14^2) = 15.64 m/s
    and
    tan theta = 14/6.99
    theta = 63.5 degrees

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    posted by Damon

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